怼就完事了，总结几种验证码的解决方案

import tesserocr
from PIL import Image
image = Image.open('code2.jpg')
image = image.convert('L')
threshold = 127
table = []
for i in range(256):
    if i < threshold:
        table.append(0)
    else:
        table.append(1)
image = image.point(table, '1')
result = tesserocr.image_to_text(image)
print(result)

# -*- coding: utf-8 -*-
import cv2
import numpy as np

imagepath = '9_1.png'
img = cv2.imread(imagepath)
gray = cv2.cvtColor ( img , cv2.COLOR_BGR2GRAY )
ret, binary = cv2.threshold(gray,127,255,cv2.THRESH_BINARY)  

contours, hierarchy = cv2.findContours(binary,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)  
#cv2.drawContours(img,contours,-1,(0,0,255),1)  
for cnt in contours:

    # 最小外界矩形的宽度和高度
    width, height = cv2.minAreaRect(cnt)[1]

    if width* height > 100:
        # 最小的外接矩形
        rect = cv2.minAreaRect(cnt)
        box = cv2.boxPoints(rect)  # 获取最小外接矩形的4个顶点

        box = np.int0(box)
        print box

        if 0 not in box.ravel():

            #绘制最小外界矩形
            for i in range(4):
                cv2.line(img, tuple(box[i]), tuple(box[(i+1)%4]), 0)  # 5
            theta = cv2.minAreaRect(cnt)[2]
            if abs(theta) <= 45:
                print('图片的旋转角度为%s.'%theta)


            #     angle = theta
print theta            
cv2.imshow("img", img)  
cv2.waitKey(0)  

import numpy as np
import math


def ease_out_expo(x):
    """
    曲线函数
    :param x:
    :return:
    """
    if x == 1:
        return 1
    else:
        return 1 - pow(2, -10 * x)


def get_tracks(distance, seconds):
    """
    轨迹生成函数
    :param distance: 滑动总距离
    :param seconds: 滑动总时间
    :return:
    """
    tracks = [0]  # 存放轨迹的数组
    offsets = [0]  # 存放滑动总距离的记录数组
    for t in np.arange(0.0, seconds, 0.1):  # 产生一个数列如[0.0, 0.1, 0.2, 0.3]
        offset = round(ease_out_expo(t/seconds) * distance)  # 根据时间t计算在曲线上的滑动距离
        tracks.append(offset - offsets[-1])  # 本次计算的距离减去上一次移动的距离，得到本次的轨迹
        offsets.append(offset)  # 至本次滑动了的总距离
    return offsets, tracks


a, b = get_tracks(138, 3)
print(a, b)


def get_tracksb(distance):
    """
    根据物理的先加速再减速规律计算
    :param distance:
    :return:
    """
    distance += 20  # 加上20是为了滑动超过缺口再回滑
    v = 0  # 初速度
    t = 0.2  # 以0.2秒为一个计算周期
    forward_tracks = []  # 轨迹记录数组
    current = 0  # 初始移动距离
    mid = distance * 3 / 5  # 减速阀值即五分之三的距离加速剩下距离减速
    while current < distance:  # 总移动距离等于输入距离时结束
        if current < mid:  # 加速状态
            a = 2  # 加速度为+2
        else:  # 减速状态
            a = -3  # 加速度-3

        s = v * t + 0.5 * a * (t ** 2)  # 计算0.2秒周期内的位移
        v = v + a * t  # 计算本次周期后的速度
        current += s  # 将之前移动的总距离，加上本次0.2秒周期内移动的距离
        forward_tracks.append(round(s))  # 记录本次0.2秒周期内的移动距离为轨迹

    back_tracks = [-3, -3, -2, -2, -2, -2, -2, -1, -1, -1]  # 手动将开头加上的20，生成减去轨迹，即回滑轨迹
    return {'forward_tracks': forward_tracks, 'back_tracks': back_tracks}

https://2captcha.com/in.php
?key= 上面获取的API KEY 
&method=geetest
&gt= 某验参数
&challenge= 某验参数
&api_server=api-na.geetest.com(可选)
&pageurl= 滑动验证码所在的网页地址

OK|2122988149 or as JSON {"status":1,"request":"2122988149"}

https://2captcha.com/res.php
?key=API KEY
&action=get
&id=2122988149

{
      "challenge":"1a2b3456cd67890e12345fab678901c2de",
      "validate":"09fe8d7c6ba54f32e1dcb0a9fedc8765",
      "seccode":"12fe3d4c56789ba01f2e345d6789c012|jordan"
}